हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा १२

A random variable X has the following probability mass function. x 1 2 3 4 5 F(x) k2 2k2 3k2 2k 3k Find P(X > 3)

Advertisements
Advertisements

प्रश्न

A random variable X has the following probability mass function.

x 1 2 3 4 5
F(x) k2 2k2 3k2 2k 3k

Find P(X > 3)

योग
Advertisements

उत्तर

P(X > 3) = P(X = 4) + P(X = 5)

= `2/6 + 3/6`

= `5/6`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Probability Distributions - Exercise 11.2 [पृष्ठ १९४]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 11 Probability Distributions
Exercise 11.2 | Q 6. (iii) | पृष्ठ १९४

संबंधित प्रश्न

Suppose error involved in making a certain measurement is continuous r.v. X with p.d.f.

f (x) = k `(4 – x^2 )`, for –2 ≤ x ≤ 2 and = 0 otherwise.

P(x > 0)


Suppose error involved in making a certain measurement is continuous r.v. X with p.d.f.

`"f(x)" = {("k"(4 - x^2)      "for –2 ≤ x ≤ 2,"),(0                                 "otherwise".):}`

P(–1 < x < 1)


Given the p.d.f. of a continuous r.v. X ,

f (x) = `x^2 /3` , for –1 < x < 2 and = 0 otherwise

Determine c.d.f. of X hence find P( x < –2)


The p.m.f. of a r.v. X is given by P (X = x) =`("" ^5 C_x ) /2^5` , for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise.

Then show that P (X ≤ 2) = P (X ≥ 3).


F(x) is c.d.f. of discrete r.v. X whose p.m.f. is given by P(x) = `"k"^4C_x` , for x = 0, 1, 2, 3, 4 and P(x) = 0 otherwise then F(5) = _______


c.d.f. of a discrete random variable X is


The probability distribution of a r.v. X is

X = x -3 -2 -1 0 1
P(X = x) 0.3 0.2 0.25 0.1 0.15

Then F (-1) = ?


The cumulative distribution function of a discrete random variable is given by
F(x) = `{{:(0,  - oo < x < - 1),(0.15, - 1 ≤ x < 0),(0.35, 0 ≤ x < 1),(0.60, 1 ≤ x < 2),(0.85, 2 ≤ x < 3),(1, 3 ≤ x < oo):}`
Find P(X < 1)


The cumulative distribution function of a discrete random variable is given by
F(x) = `{{:(0,  "for" - oo < x < 0),(1/2,  "for"  0 ≤ x < 1),(3/5,  "for"  1 ≤ x < 2),(4/5,  "for"  2 ≤ x < 4),(9/5,  "for"  3 ≤ x < 4),(1,  "for"   ≤ x < oo):}`
Find the probability mass function


If Xis a.r.v. with c.d.f F (x) and its probability distribution is given by

X = x - 1.5 -0.5 0.5 1.5 2.5
P(X = x) 0.05 0.2 0.15 0.25 0.35

then, F(1.5) - F(- 0.5) = ?


Choose the correct alternative:

Two coins are to be flipped. The first coin will land on heads with probability 0.6, the second with Probability 0.5. Assume that the results of the flips are independent and let X equal the total number of heads that result. The value of E[X] is


Let X = time (in minutes) that lapses between the ringing of the bell at the end of a lecture and the actual time when the professor ends the lecture. Suppose X has p.d.f.

f(x) = `{(kx^2","      0 ≤ x ≤ 2), (0","         "othenwise"):}`

Then, the probability that the lecture ends within 1 minute of the bell ringing is ______


If A = {x ∈ R : x2 - 5 |x| + 6 = 0}, then n(A) = _____.


If the probability function of a random variable X is defined by P(X = k) = a`((k + 1)/2^k)` for k - 0, 1, 2, 3, 4, 5, then the probability that X takes a prime value is ______


The probability distribution of a random variable X is given below.

X = k 0 1 2 3 4
P(X = k) 0.1 0.4 0.3 0.2 0

The variance of X is ______


Two coins are tossed. Then the probability distribution of number of tails is.


At random variable X – B(n, p), if values of mean and variance of X are 18 and 12 respectively, then total number of possible values of X are ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×